Giải phương trình sau $\frac{4}{2x-3}$ + $\frac{x}{4x^{2}-1}$ = $\frac{1}{2x+3}$

Giải thích các bước giải:

ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne  \pm \frac{3}{2}\\
x \ne  \pm \frac{1}{2}
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
\frac{4}{{2x – 3}} + \frac{x}{{4{x^2} – 1}} = \frac{1}{{2x + 3}}\\
 \Leftrightarrow \frac{x}{{4{x^2} – 1}} = \frac{1}{{2x + 3}} – \frac{4}{{2x – 3}}\\
 \Leftrightarrow \frac{x}{{4{x^2} – 1}} = \frac{{4\left( {2x – 3} \right) – \left( {2x + 3} \right)}}{{4\left( {2x + 3} \right)\left( {2x – 3} \right)}}\\
 \Leftrightarrow \frac{x}{{4{x^2} – 1}} = \frac{{6x – 15}}{{4\left( {4{x^2} – 9} \right)}}\\
 \Leftrightarrow 4x\left( {4{x^2} – 9} \right) = \left( {4{x^2} – 1} \right)\left( {6x – 15} \right)\\
 \Leftrightarrow 16{x^3} – 36x = 24{x^3} – 60{x^2} – 6x + 15\\
 \Leftrightarrow 8{x^3} – 60{x^2} + 30x + 15 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 6,91\\
x = 0,89\\
x =  – 0,3
\end{array} \right.
\end{array}\)